Question

An object is thrown upward from the top of a 64 -foot building with an initial velocity of 48 feet per second. (a) Write the position, velocity, and acceleration functions of the object. (b) When will the object hit the ground? (c) When is the velocity of the object zero? (d) How high does the object go? (e) Use a graphing utility to graph the position, velocity, and acceleration functions in the same viewing window. Write a short paragraph that describes the relationship among these functions.

Short answer

The position, velocity, and acceleration functions are \( h(t) = -16t^2 + 48t + 64 \), \( v(t) = -32t + 48 \), and \( a(t) = -32 \) respectively. The object hits the ground at 4 seconds, velocity becomes zero at 1.5 seconds, and the maximum height it reaches is 100 feet.

Step-by-step solution

Construct the Position Function

Using the formula for the height of an object in free fall, \( h(t) = -16t^2 + v_0t + h_0 \), where \( h_0 \) is the initial height, \( v_0 \) is the initial velocity, and \( t \) is the time. Plugging in given values, \( h(t) = -16t^2 + 48t + 64 \).

Construct the Velocity Function

The velocity function can be found by differentiating the position function, yielding \( v(t) = -32t + 48 \).

Construct the Acceleration Function

The acceleration function, which is the derivative of the velocity function or the second derivative of the position function, for an object in free fall is a constant, \( a(t) = -32 \).

Determine When the Object Hits the Ground

This can be found by setting the position function equal to zero and solving for \( t \). The positive solution is the time it takes for the object to hit the ground, so \( -16t^2 + 48t + 64 = 0 \). Solving this quadratic equation gives \( t = 4 \) seconds.

Determine When Velocity is Zero

Set the velocity function equal to zero and solve for \( t \) yielding \( -32t + 48 = 0 \). We find that \( t = 1.5 \) seconds.

Determine Maximum Height

The object reaches its maximum height when its velocity is zero. So plugging \( t = 1.5 \) into the position equation gives the maximum height of \( h(1.5) = 100 \) feet.

Graph and Interpret Functions

This step requires a graphing utility, making sure to graph all three functions on the same axes. The position function is a downward-opening parabola, representing the upward and downward trajectory of the object, reaching maximum at \( t = 1.5 \). The velocity function is a straight line, crossing the x-axis at \( t = 1.5 \), representing the point of zero velocity. The acceleration function is a horizontal line at \( -32 \), indicating the constant acceleration due to gravity.

Key concepts

Quadratic Functions

Quadratic functions are polynomial functions of degree two and can be written in the general form of \( f(x) = ax^2 + bx + c \), where \(a\), \(b\), and \(c\) are constants and \(a \eq 0\). They are characterized by their distinct 'U' shaped curve called a parabola. In the context of free fall motion equations, quadratic functions come into play when expressing the position of an object as a function of time.

The position function in our exercise \( h(t) = -16t^2 + 48t + 64 \) is a quadratic function where the coefficient of \(t^2\) is negative, indicating a downward-opening parabola. This shape is crucial as it represents the object's ascent up to a peak point (the vertex), and subsequent descent, just like the path an object takes when thrown upward against gravity. The highest point on this graph (the vertex) can be found using the formula \( -b/(2a) \) and represents the point at which the thrown object reaches its maximum height.

Derivative Applications

The application of derivatives is a fundamental concept in calculus, particularly when analyzing motion. In physics, the first derivative of the position with respect to time represents the velocity of an object, while the second derivative represents its acceleration.

In our example, the first derivative \( v(t) = -32t + 48 \) of the position function gives the velocity of the object. When we calculate the velocity function, we are finding the rate of change of height with respect to time. Setting this function to zero tells us when the object's motion changes direction—that is, when it has zero velocity and reaches its maximum height, which occurs at \( t = 1.5 \) seconds in this exercise. On the other hand, acceleration, derived here as a constant \( a(t) = -32 \) by differentiating the velocity function or by taking the second derivative of the position function, describes the rate at which the object's velocity changes over time. In the context of our planet, it represents the acceleration due to gravity (in feet per second squared).

Graphical Analysis of Motion

Graphing the functions related to an object's motion provides a clear visualization of how an object moves over time. The position, velocity, and acceleration graphs yield insight into the behavior of the object as it follows its trajectory.

The position graph, a parabola for free-falling objects, showcases how the height changes over time. From the exercise, plotting \( h(t) \) reveals a downward-opening curve peaking at 1.5 seconds—this is where the object attains maximum height before falling back down. The velocity graph is a straight line that intersects the time axis at the point where the object comes to a momentary stop (zero velocity) before reversing direction. Lastly, the acceleration graph is a horizontal line, reflecting the unchanging acceleration due to gravity, which remains constant regardless of the object's position or speed. By seeing these graphs together, students can understand the dependency and interaction between position, velocity, and acceleration in the object's motion.

Calculus in Physics

Calculus plays a pivotal role in physics, especially when analyzing dynamics such as motion. The differentiation and integration processes of calculus allow us to move between position, velocity, and acceleration functions.

As shown in the exercise, calculus enables us to derive the velocity function by differentiating the position function and the acceleration by differentiating the velocity function, or by taking the second derivative of the position function. The constant acceleration of gravity illustrates that the force exerted on falling objects is consistent, influencing their velocity and hence their position over time. Through calculus, complex physical phenomena can be expressed in precise mathematical terms, and we can predict an object's future behavior by understanding its current state. Notably, calculus not only helps to determine when the object will reach the ground (\( t = 4 \) seconds), but also to reason about the instantaneous changes in motion, underscoring its indispensable role in physics and engineering.